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 Gone Fishin' (Posted on 2005-12-05)
A man is sitting in a lake in his boat fishing when he receives a call on his cell phone. A barbecue is happening a ways down the shoreline, and he had better get there fast so as not to miss out. He is two miles out perpendicular to the shore, and 7 miles horizontally from the location on the beach from the barbecue. If the lake has no current and the wind is negligible, he can row toward the shoreline at a rate of 3 mph. When he reaches dry land, he can run at 5 miles per hour. If he wants to reach the barbecue as quickly as possible, how far horizontally should he land the boat from his current location?

As a bonus, if we assign the distance from the shore to be A miles, the distance from the barbecue along the shoreline B miles, and the boat speed and running speed C and D miles per hour respectively, does there exist a function that will output the ideal place to land the boat for all positive values of A,B,C, and D? If so, what is it? If not, why not?

 See The Solution Submitted by Dan Rating: 3.6667 (3 votes)

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 Solution | Comment 1 of 5
` `
`Let's give cartesian coordinates to the boat and barbecue of (0,A) and(B,0) respectively. Clearly the man wants to row to the point (x,0)where 0 <= x <= B. The time required to get to the barbecue is given by`
`          sqrt(A^2 + x^2)     B - x     T = ----------------- + -------                C               D`
`   dT          x                1  ---- = ------------------- - ---   dx     C sqrt(A^2 + x^2)     D`
`Setting dT/dx to zero an solving for x gives`
`               A*C     x = -----------------          sqrt(D^2 - C^2)`
`Plugging in the values for our problem gives`
`               2*3            3     x = ----------------- = ---          sqrt(5^2 - 3^2)     2`
`As for the bonus. If C >= D , then we have no solution. If C < D ,then we still require`
`            A*C     ----------------- <= B      sqrt(D^2 - C^2) `

 Posted by Bractals on 2005-12-05 09:47:03

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